All real and imaginary numbers and multiplicity for x^4+2x^2+1=0. show work.

x = i, mult 2; x = -i, mult 2

Step-by-step explanation:

First let's make a substitution to make this easier to factor.  

Let u² = x⁴ and

      u = x²

Now we can rewrite the polynomial as

u² +2u + 1 = 0

This factors easily into

(u + 1)(u + 1) = 0

By the Zero Product Property, either

u + 1 = 0 or u + 1 = 0

Putting back the x²:

x² + 1 = 0 or x² + 1 = 0

For the first one, even though they are the same:

x² = -1 so

x = ±√-1

Since that is not "allowed", we make the replacement of -1 = i²:

x = ±√i² so

x = ±i

For the second one, we need not repeat the whole process, but we find 2 more identical roots:

x = ±i

That means that the factors are

(x + i)(x - i)(x + i)(x - i) = 0

x = i, multiplicity 2 and

x = -i, multiplicity 2

step-by-step explanation:

the true ones are a b and d